Answer:
Step-by-step explanation:
Transformation #1 (blue grafic): rotation in y axis
Transformation #2 (green grafic): translation in y axis.
I hope I've helped you.
I hope you've understand me.
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1 answer:
Answer:
1. 7x+5y=19
(+)-7x-2y=-16 step one
____________
=
step two
y=1
____________
since you have the value of y u can plug it in in either of the equations.
7x+5(1)=19 step three
____________
7x+5-5=19-5 step four
____________
=
x=2 step five
____________
You can check whether the answers are correct by plugging them back in.
7(2)+5(1)=19
14+5=19
19 equals 19 so it checks.
You can do the same thing for the rest.
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Answer:
Step-by-step explanation:
Given
Required
Determine the roots
To get the root of the function, then f(b) must be 0;
i.e. f(b) = 0
So, the expression becomes
Add 75 to both sides
Take square roots of both sides
Reorder
Expand 75 as a product of 25 and 3
Split the expression
The options are not clear enough; however the roots of the equation are
So if it is 60 miles per hour it is basically 60 miles per 60 minutes. when you simplify it, it is going to be 1/1=1 The answer is 1 mile per minute
Answer:
Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))
1. R is reflexive if for each element a∈B, aRa.
2. R is symmetric if satisfies that if aRb then bRa.
3. R is transitive if satisfies that if aRb and bRc then aRc.
Then, our set B is .
a) We need to find a relation R reflexive and transitive that contain the relation
Then, we need:
1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,
2. Observe that
- 1R4 and 4R1, then 1 must be related with itself.
- 4R1 and 1R4, then 4 must be related with itself.
- 4R1 and 1R2, then 4 must be related with 2.
Therefore is the smallest relation containing the relation R1.
b) We need a new relation symmetric and transitive, then
- since 1R2, then 2 must be related with 1.
- since 1R4, 4 must be related with 1.
and the analysis for be transitive is the same that we did in a).
Observe that
- 1R2 and 2R1, then 1 must be related with itself.
- 4R1 and 1R4, then 4 must be related with itself.
- 2R1 and 1R4, then 2 must be related with 4.
- 4R1 and 1R2, then 4 must be related with 2.
- 2R4 and 4R2, then 2 must be related with itself
Therefore, the smallest relation containing R1 that is symmetric and transitive is
c) We need a new relation reflexive, symmetric and transitive containing R1.
For be reflexive
- 1 must be related with 1,
- 2 must be related with 2,
- 3 must be related with 3,
- 4 must be related with 4
For be symmetric
- since 1R2, 2 must be related with 1,
- since 1R4, 4 must be related with 1.
For be transitive
- Since 4R1 and 1R2, 4 must be related with 2,
- since 2R1 and 1R4, 2 must be related with 4.
Then, the smallest relation reflexive, symmetric and transitive containing R1 is